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Orbital Mechanics for
Engineering Students
To my parents, Rondo and Geraldine, and my wife, Connie Dee
Orbital Mechanics for
Engineering Students
Howard D. Curtis
Embry-Riddle Aeronautical University
Daytona Beach, Florida
AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD
PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO
Elsevier Butterworth-Heinemann
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First published 2005
Copyright © 2005, Howard D. Curtis. All rights reserved
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Contents
Preface xi
Supplements to the text xv
Chapter
1 Dynamics of point masses 1
1.1 Introduction 1
1.2 Kinematics 2
1.3 Mass, force and Newton’s law of gravitation 7
1.4 Newton’s law of motion 10
1.5 Time derivatives of moving vectors 15
1.6 Relative motion 20
Problems 29
Chapter
2 The two-body problem 33
2.1 Introduction 33
2.2 Equations of motion in an inertial frame 34
2.3 Equations of relative motion 37
2.4 Angular momentum and the orbit formulas 42
2.5 The energy law 50
2.6 Circular orbits (e = 0) 51
2.7 Elliptical orbits (0 < e < 1) 55
2.8 Parabolic trajectories (e = 1) 65
2.9 Hyperbolic trajectories (e > 1) 69
2.10 Perifocal frame 76
2.11 The Lagrange coefficients 78
2.12 Restricted three-body problem 89
2.12.1 Lagrange points 92
2.12.2 Jacobi constant 96
Problems 101
Chapter
3 Orbital position as a function of time 107
3.1 Introduction 107
3.2 Time since periapsis 108
v
vi Contents
3.3 Circular orbits 108
3.4 Elliptical orbits 109
3.5 Parabolic trajectories 124
3.6 Hyperbolic trajectories 125
3.7 Universal variables 134
Problems 145
Chapter
4 Orbits in three dimensions 149
4.1 Introduction 149
4.2 Geocentric right ascension–declination frame 150
4.3 State vector and the geocentric equatorial frame 154
4.4 Orbital elements and the state vector 158
4.5 Coordinate transformation 164
4.6 Transformation between geocentric equatorial and
perifocal frames 172
4.7 Effects of the earth’s oblateness 177
Problems 187
Chapter
5 Preliminary orbit determination 193
5.1 Introduction 193
5.2 Gibbs’ method of orbit determination from three
position vectors 194
5.3 Lambert’s problem 202
5.4 Sidereal time 213
5.5 Topocentric coordinate system 218
5.6 Topocentric equatorial coordinate system 221
5.7 Topocentric horizon coordinate system 223
5.8 Orbit determination from angle and range
measurements 228
5.9 Angles-only preliminary orbit determination 235
5.10 Gauss’s method of preliminary orbit determination 236
Problems 250
Chapter
6 Orbital maneuvers 255
6.1 Introduction 255
6.2 Impulsive maneuvers 256
6.3 Hohmann transfer 257
Contents vii
6.4 Bi-elliptic Hohmann transfer 264
6.5 Phasing maneuvers 268
6.6 Non-Hohmann transfers with a common apse line 273
6.7 Apse line rotation 279
6.8 Chase maneuvers 285
6.9 Plane change maneuvers 290
Problems 304
Chapter
7 Relative motion and rendezvous 315
7.1 Introduction 315
7.2 Relative motion in orbit 316
7.3 Linearization of the equations of relative motion in
orbit 322
7.4 Clohessy–Wiltshire equations 324
7.5 Two-impulse rendezvous maneuvers 330
7.6 Relative motion in close-proximity circular orbits 338
Problems 340
Chapter
8 Interplanetary trajectories 347
8.1 Introduction 347
8.2 Interplanetary Hohmann transfers 348
8.3 Rendezvous opportunities 349
8.4 Sphere of influence 354
8.5 Method of patched conics 359
8.6 Planetary departure 360
8.7 Sensitivity analysis 366
8.8 Planetary rendezvous 368
8.9 Planetary flyby 375
8.10 Planetary ephemeris 387
8.11 Non-Hohmann interplanetary trajectories 391
Problems 398
Chapter
9 Rigid-body dynamics 399
9.1 Introduction 399
9.2 Kinematics 400
9.3 Equations of translational motion 408
9.4 Equations of rotational motion 410
viii Contents
9.5 Moments of inertia 414
9.5.1 Parallel axis theorem 428
9.6 Euler’s equations 435
9.7 Kinetic energy 441
9.8 The spinning top 443
9.9 Euler angles 448
9.10 Yaw, pitch and roll angles 459
Problems 463
Chapter
10 Satellite attitude dynamics 475
10.1 Introduction 475
10.2 Torque-free motion 476
10.3 Stability of torque-free motion 486
10.4 Dual-spin spacecraft 491
10.5 Nutation damper 495
10.6 Coning maneuver 503
10.7 Attitude control thrusters 506
10.8 Yo-yo despin mechanism 509
10.9 Gyroscopic attitude control 516
10.10 Gravity-gradient stabilization 530
Problems 543
Chapter
11 Rocket vehicle dynamics 551
11.1 Introduction 551
11.2 Equations of motion 552
11.3 The thrust equation 555
11.4 Rocket performance 557
11.5 Restricted staging in field-free space 560
11.6 Optimal staging 570
11.6.1 Lagrange multiplier 570
Problems 578
References and further reading 581
Appendix
A Physical data 583
Appendix
B A road map 585
Contents ix
Appendix
C Numerical integration of the n-body
equations of motion 587
C.1 Function file accel_3body.m 590
C.2 Script file threebody.m 592
Appendix
D MATLAB algorithms 595
D.1 Introduction 596
D.2 Algorithm 3.1: solution of Kepler’s equation by
Newton’s method 596
D.3 Algorithm 3.2: solution of Kepler’s equation for the
hyperbola using Newton’s method 598
D.4 Calculation of the Stumpff functions S(z) and C(z) 600
D.5 Algorithm 3.3: solution of the universal Kepler’s
equation using Newton’s method 601
D.6 Calculation of the Lagrange coefficients f and g and
their time derivatives 603
D.7 Algorithm 3.4: calculation of the state vector (r, v)
given the initial state vector (r0, v0) and the
time lapse �t 604
D.8 Algorithm 4.1: calculation of the orbital elements from
the state vector 606
D.9 Algorithm 4.2: calculation of the state vector from
the orbital elements 610
D.10 Algorithm 5.1: Gibbs’ method of preliminary orbit
determination 613
D.11 Algorithm 5.2: solution of Lambert’s problem 616
D.12 Calculation of Julian day number at 0 hr UT 621
D.13 Algorithm 5.3: calculation of local sidereal time 623
D.14 Algorithm 5.4: calculation of the state vector
from measurements of range, angular position and
their rates 626
D.15 Algorithms 5.5 and 5.6: Gauss’s method of preliminary
orbit determination with iterative improvement 631
D.16 Converting the numerical designation of a month or
a planet into its name 640
D.17 Algorithm 8.1: calculation of the state vector of
a planet at a given epoch 641
D.18 Algorithm 8.2: calculation of the spacecraft trajectory
from planet 1 to planet 2 648
Appendix
E Gravitational potential energy of a sphere657
Index 661
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Preface
This textbook evolved from a formal set of notes developed over nearly ten years
of teaching an introductory course in orbital mechanics for aerospace engineering
students. These undergraduate students had no prior formal experience in the subject,
but had completed courses in physics, dynamics and mathematics through differential
equations and applied linear algebra. That is the background I have presumed for
readers of this book.
This is by no means a grand, descriptive survey of the entire subject of astronautics.
It is a foundations text, a springboard to advanced study of the subject. I focus on the
physical phenomena and analytical procedures required to understand and predict, to
first order, the behavior of orbiting spacecraft. I have tried to make the book readable
for undergraduates, and in so doing I do not shy away from rigor where it is needed
for understanding. Spacecraft operations that take place in earth orbit are considered
as are interplanetary missions. The important topic of spacecraft control systems is
omitted. However, the material in this book and a course in control theory provide
the basis for the study of spacecraft attitude control.
A brief perusal of the Contents shows that there are more than enough topics
to cover in a single semester or term. Chapter 1 is a review of vector kinematics in
three dimensions and of Newton’s laws of motion and gravitation. It also focuses on
the issue of relative motion, crucial to the topics of rendezvous and satellite attitude
dynamics. Chapter 2 presents the vector-based solution of the classical two-body
problem, coming up with a host of practical formulas for orbit and trajectory analy-
sis. The restricted three-body problem is covered in order to introduce the notion of
Lagrange points. Chapter 3 derives Kepler’s equations, which relate position to time
for the different kinds of orbits. The concept of ‘universal variables’ is introduced.
Chapter 4 is devoted to describing orbits in three dimensions and accounting for the
major effects of the earth’s oblate, non-spherical shape. Chapter 5 is an introduction
to preliminary orbit determination, including Gibbs’ and Gauss’s methods and the
solution of Lambert’s problem. Auxiliary topics include topocentric coordinate sys-
tems, Julian day numbering and sidereal time. Chapter 6 presents the common means
of transferring from one orbit to another by impulsive delta-v maneuvers, including
Hohmann transfers, phasing orbits and plane changes. Chapter 7 derives and employs
the equations of relative motion required to understand and design two-impulse ren-
dezvous maneuvers. Chapter 8 explores the basics of interplanetary mission analysis.
Chapter 9 presents those elements of rigid-body dynamics required to characterize
the attitude of an orbiting satellite. Chapter 10 describes the methods of controlling,
changing and stabilizing the attitude of spacecraft by means of thrusters, gyros and
other devices. Finally, Chapter 11 is a brief introduction to the characteristics and
design of multi-stage launch vehicles.
Chapters 1 through 4 form the core of a first orbital mechanics course. The time
devoted to Chapter 1 depends on the background of the student. It might be surveyed
xi
xii Preface
briefly and used thereafter simply as a reference. What follows Chapter 4 depends on
the objectives of the course.
Chapters 5 through 8 carry on with the subject of orbital mechanics. Chapter 6
on orbital maneuvers should be included in any case. Coverage of Chapters 5, 7 and
8 is optional. However, if all of Chapter 8 on interplanetary missions is to form a part
of the course, then the solution of Lambert’s problem (Section 5.3) must be studied
beforehand.
Chapters 9 and 10 must be covered if the course objectives include an introduction
to satellite dynamics. In that case Chapters 5, 7 and 8 would probably not be studied
in depth.
Chapter 11 is optional if the engineering curriculum requires a separate course in
propulsion, including rocket dynamics.
To understand the material and to solve problems requires using a lot of under-
graduate mathematics. Mathematics, of course, is the language of engineering.
Students must not forget that Sir Isaac Newton had to invent calculus so he could solve
orbital mechanics problems precisely. Newton (1642–1727) was an English physi-
cist and mathematician, whose 1687 publication Mathematical Principles of Natural
Philosophy (‘the Principia’) is one of the most influential scientific works of all time. It
must be noted that the German mathematician Gottfried Wilhelm von Leibniz (1646–
1716) is credited with inventing infinitesimal calculus independently of Newton in
the 1670s.
In addition to honing their math skills, students are urged to take advantage
of computers (which, incidentally, use the binary numeral system developed by
Leibniz). There are many commercially available mathematics software packages for
personal computers. Wherever possible they should be used to relieve the burden of
repetitive and tedious calculations. Computer programming skills can and should be
put to good use in the study of orbital mechanics. Elementary MATLAB® programs
(M-files) appear at the end of this book to illustrate how some of the procedures devel-
oped in the text can be implemented in software. All of the scripts were developed
using MATLAB version 5.0 and were successfully tested using version 6.5 (release 13).
Information about MATLAB, which is a registered trademark of The MathWorks,
Inc., may be obtained from:
The MathWorks, Inc.
3 Apple Hill Drive
Natick, MA, 01760-2098 USA
Tel: 508-647-7000
Fax: 508-647-7101
E-mail: info@mathworks.com
Web: www.mathworks.com
The text contains many detailed explanations and worked-out examples. Their
purpose is not to overwhelm but to elucidate. It is always assumed that the material is
being seen for the first time and, wherever possible, solution details are provided so as
to leave little to the reader’s imagination. There are some exceptions to this objective,
deemed necessary to maintain the focus and control the size of the book. For example,
in Chapter 6, the notion of specific impulse is laid on the table as a means of rating
rocket motor performance and to show precisely how delta-v is related to propellant
expenditure. In Chapter 10 Routh–Hurwitz stability criteria are used without proof to
Preface xiii
show quantitatively that a particular satellite configuration is, indeed, stable. Specific
impulse is covered in more detail in Chapter 11, and the stability of linear systems is
treated in depth in books on control theory. See, for example, Nise (2003) and Ogata
(2001).
Supplementary material appears in the appendices at the end of the book.
Appendix A lists physical data for use throughout the text. Appendix B is a ‘road
map’ to guide the reader through Chapters 1, 2 and 3. Appendix C shows how to set
up the n-body equations of motion and program them in MATLAB. Appendix D lists
the MATLAB implementations of algorithms presented in several of the chapters.
Appendix E shows that the gravitational field of a spherically symmetric body is the
same as if the mass were concentrated at its center.
The field of astronautics is rich and vast. References cited throughout this text are
listed at the end of the book. Also listed are other books on the subject that might be
of interest to those seeking additional insights.
I wish to thank colleagues who provided helpful criticism and advice during the
development of this book. Yechiel Crispin and Charles Eastlake were sources for
ideas about what should appear in the summary chapter on rocket dynamics. Habib
Eslami, Lakshmanan Narayanaswami, Mahmut Reyhanoglu and Axel Rohde all used
the evolving manuscript as either a text or a reference in their space mechanics courses.
Based on their classroom experiences, they gave me valuable feedback in the form
of corrections, recommendations and much-needed encouragement.Tony Hagar
voluntarily and thoroughly reviewed the entire manuscript and made a number of
suggestions, nearly all of which were incorporated into the final version of the text.
I am indebted to those who reviewed the manuscript for the publisher for their
many suggestions on how the book could be improved and what additional topics
might be included.
Finally, let me acknowledge how especially grateful I am to the students who,
throughout the evolution of the book, reported they found it to be a helpful and
understandable introduction to space mechanics.
Howard D. Curtis
Embry-Riddle Aeronautical University
Daytona Beach, Florida
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Supplements
to the text
For the student:
• Copies of the MATLAB programs (M-files) that appear in Appendix D can
be downloaded from the companion website accompanying this book. To
access these please visit http://books.elsevier.com/companions and follow the
instructions on screen.
For the instructor:
• A full Instructor’s Solutions Manual is available for adopting tutors, which pro-
vides complete worked-out solutions to the problems set at the end of each
chapter. To access these please visit http://books.elsevier.com/manuals and follow
the instructions on screen.
xv
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1C h a p t e r
Dynamics of
point masses
Chapter outline
1.1 Introduction 1
1.2 Kinematics 2
1.3 Mass, force and Newton’s law of gravitation 7
1.4 Newton’s law of motion 10
1.5 Time derivatives of moving vectors 15
1.6 Relative motion 20
Problems 29
1.1 Introduction
This chapter serves as a self-contained reference on the kinematics and dynamicsof point masses as well as some basic vector operations. The notation and
concepts summarized here will be used in the following chapters. Those familiar with
the vector-based dynamics of particles can simply page through the chapter and then
refer back to it later as necessary. Those who need a bit more in the way of review
will find the chapter contains all of the material they need in order to follow the
development of orbital mechanics topics in the upcoming chapters.
We begin with the problem of describing the curvilinear motion of particles
in three dimensions. The concepts of force and mass are considered next, along
with Newton’s inverse-square law of gravitation. This is followed by a presentation
1
2 Chapter 1 Dynamics of point masses
of Newton’s second law of motion (‘force equals mass times acceleration’) and the
important concept of angular momentum.
As a prelude to describing motion relative to moving frames of reference, we
develop formulas for calculating the time derivatives of moving vectors. These are
applied to the computation of relative velocity and acceleration. Example problems
illustrate the use of these results as does a detailed consideration of how the earth’s
rotation and curvature influence our measurements of velocity and acceleration. This
brings in the curious concept of Coriolis force. Embedded in exercises at the end of
the chapter is practice in verifying several fundamental vector identities that will be
employed frequently throughout the book.
1.2 Kinematics
To track the motion of a particle P through Euclidean space we need a frame of
reference, consisting of a clock and a cartesian coordinate system. The clock keeps
track of time t and the xyz axes of the cartesian coordinate system are used to locate
the spatial position of the particle. In non-relativistic mechanics, a single ‘universal’
clock serves for all possible cartesian coordinate systems. So when we refer to a frame
of reference we need think only of the mutually orthogonal axes themselves.
The unit of time used throughout this book is the second (s). The unit of length
is the meter (m), but the kilometer (km) will be the length unit of choice when large
distances and velocities are involved. Conversion factors between kilometers, miles
and nautical miles are listed in Table A.3.
Given a frame of reference, the position of the particle P at a time t is defined
by the position vector r(t) extending from the origin O of the frame out to P itself,
as illustrated in Figure 1.1. (Vectors will always be indicated by boldface type.) The
x
y
z
O
�
v
a
P
s
o
Pat
h
Figure 1.1 Position, velocity and acceleration vectors.
1.2 Kinematics 3
components of r(t) are just the x, y and z coordinates,
r(t) = x(t)î + y(t)ĵ + z(t)k̂
î, ĵ and k̂ are the unit vectors which point in the positive direction of the x, y and z
axes, respectively. Any vector written with the overhead hat (e.g., â) is to be considered
a vector of unit dimensionless magnitude.
The distance of P from the origin is the magnitude or length of r, denoted ‖r‖ or
just r,
‖r‖ = r =
√
x2 + y2 + z2
The magnitude of r, or any vector A for that matter, can also be computed by means
of the dot product operation,
r = √r · r ‖A‖ = √A · A
The velocity v and acceleration a of the particle are the first and second time derivatives
of the position vector,
v(t) = dx(t)
dt
î + dy(t)
dt
ĵ + dz(t)
dt
k̂ = vx(t)î + vy(t)ĵ + vz(t)k̂
a(t) = dvx(t)
dt
î + dvy(t)
dt
ĵ + dvz(t)
dt
k̂ = ax(t)î + ay(t)ĵ + az(t)k̂
It is convenient to represent the time derivative by means of an overhead dot. In this
shorthand notation, if ( ) is any quantity, then
(
·
) ≡ d()
dt
(
··
) ≡ d
2()
dt2
(
···
) ≡ d
3()
dt3
, etc.
Thus, for example,
v = ṙ
a = v̇ = r̈
vx = ẋ vy = ẏ vz = ż
ax = v̇x = ẍ ay = v̇y = ÿ az = v̇z = z̈
The locus of points that a particle occupies as it moves through space is called its path
or trajectory. If the path is a straight line, then the motion is rectilinear. Otherwise, the
path is curved, and the motion is called curvilinear. The velocity vector v is tangent
to the path. If ût is the unit vector tangent to the trajectory, then
v = vût
where v, the speed, is the magnitude of the velocity v. The distance ds that P travels
along its path in the time interval dt is obtained from the speed by
ds = v dt
louiscoo
下划线
4 Chapter 1 Dynamics of point masses
In other words,
v = ṡ
The distance s, measured along the path from some starting point, is what the odome-
ters in our automobiles record. Of course, ṡ, our speed along the road, is indicated by
the dial of the speedometer.
Note carefully that v �= ṙ, i.e., the magnitude of the derivative of r does not equal
the derivative of the magnitude of r.
Example
1.1
The position vector in meters is given as a function of time in seconds as
r = (8t2 + 7t + 6)î + (5t3 + 4)ĵ + (0.3t4 + 2t2 + 1)k̂ (m) (a)
At t = 10 seconds, calculate v (the magnitude of the derivative of r) and ṙ (the
derivative of the magnitude of r).
The velocity v is found by differentiating the given position vector with respect to
time,
v = dr
dt
= (16t + 7)î + 15t2 ĵ + (1.2t3 + 4t)k̂
The magnitude of this vector is the square root of the sum of the squares of its
components,
‖v‖ = (1.44t6 + 234.6t4 + 272t2 + 224t + 49) 12
Evaluating this at t = 10 s, we get
v = 1953.3 m/s
Calculating the magnitude of r in (a), leads to
‖r‖ = (0.09t8 + 26.2t6 + 68.6t4 + 152t3 + 149t2 + 84t + 53) 12
Differentiating this expression with respect to time,
ṙ = dr
dt
= 0.36t
7 + 78.6t5 + 137.2t3 + 228t2 + 149t + 42
(0.09t8 + 26.2t6 + 68.6t4 + 152t3 + 149t2 + 84t + 53) 12
Substituting t = 10 s, yields
ṙ = 1935.5 m/s
If v is given, then we can find the components of the unit tangent ût in the cartesian
coordinate frame of reference
ût = v‖v‖ =
vx
v
î + vy
v
ĵ + vz
v
k̂
(
v =
√
v2x + v2y + v2z
)
1.2 Kinematics 5
The acceleration may be written,
a = at ût + anûn
where at and an are the tangential and normal components of acceleration, given by
at = v̇ (= s̈) an = v
2
�
(1.1)
� is the radius of curvature, which is the distance from the particle P to the center of
curvature of the path at that point. The unit principal normal ûn is perpendicular to
ût and points towards the center of curvature C, as shown in Figure 1.2. Therefore,
theposition of C relative to P, denoted rC/P , is
rC/P = �ûn
The orthogonal unit vectors ût and ûn form a plane called the osculating plane. The
unit normal to the osculating plane is ûb, the binormal, and it is obtained from ût
and ûn by taking their cross product,
ûb = ût × ûn
The center of curvature lies in the osculating plane. When the particle P moves an
incremental distance ds the radial from the center of curvature to the path sweeps
out a small angle dφ, measured in the osculating plane. The relationship between this
angle and ds is
ds = � dφ
so that ṡ = �φ̇, or
φ̇ = v
�
(1.2)
x
y
z
O
P
C df
ds
ut
un
ub
r
Osculating plane
ˆ ˆ
ˆ
Figure 1.2 Orthogonal triad of unit vectors associated with the moving point P.
6 Chapter 1 Dynamics of point masses
Example
1.2
Relative to a cartesian coordinate system, the position, velocity and acceleration of a
particle relative at a given instant are
r = 250î + 630ĵ + 430k̂ (m)
v = 90î + 125ĵ + 170k̂ (m/s)
a = 16î + 125ĵ + 30k̂ (m/s2)
Find the coordinates of the center of curvature at that instant.
First, we calculate the speed v,
v = ‖v‖ =
√
902 + 1252 + 1702 = 229.4 m/s
The unit tangent is, therefore,
ût = v
v
= 90î + 125ĵ + 170k̂
797.4
= 0.3923î + 0.5449ĵ + 0.7411k̂
We project the acceleration vector onto the direction of the tangent to get its tangential
component at ,
at = a · ût = (16î + 125ĵ + 30k̂) · (0.3923î + 0.5449ĵ + 0.7411k̂) = 96.62 m/s2
The magnitude of a is
a =
√
162 + 1252 + 302 = 129.5 m/s2
Since a = at ût + anûn and ût and ûn are perpendicular to each other, it follows that
a2 = a2t + a2n, which means
an =
√
a2 − a2t =
√
129.52 − 96.622 = 86.29 m/s2
Hence,
ûn = 1
an
(a − at ût )
= 1
86.29
[(16î + 125ĵ + 30k̂) − 96.62(0.3923î + 0.5449ĵ + 0.7411k̂)]
= −0.2539î + 0.8385ĵ − 0.4821k̂
The equation an = v2/� can now be solved for � to yield
� = v
2
an
= 229.4
2
86.29
= 609.9 m
1.3 Mass, force and Newton’s law of gravitation 7
Let rC be the position vector of the center of curvature C. Then
rC = r + rC/P
= r + �ûn = 250î + 630ĵ + 430k̂ + 609.9(−0.2539î + 0.8385ĵ − 0.4821k̂)
= 95.16î + 1141ĵ + 136.0k̂ (m)
That is, the coordinates of C are
x = 95.16 m y = 1141 m z = 136.0 m
1.3 Mass, force and Newton’s law of
gravitation
Mass, like length and time, is a primitive physical concept: it cannot be defined in
terms of any other physical concept. Mass is simply the quantity of matter. More
practically, mass is a measure of the inertia of a body. Inertia is an object’s resistance
to changing its state of motion. The larger its inertia (the greater its mass), the more
difficult it is to set a body into motion or bring it to rest. The unit of mass is the
kilogram (kg).
Force is the action of one physical body on another, either through direct contact
or through a distance. Gravity is an example of force acting through a distance, as are
magnetism and the force between charged particles. The gravitational force between
two masses m1 and m2 having a distance r between their centers is
Fg = G m1m2
r2
(1.3)
This is Newton’s law of gravity, in which G, the universal gravitational constant, has
the value 6.6742 × 1011 m3/kg · s2. Due to the inverse-square dependence on distance,
the force of gravity rapidly diminishes with the amount of separation between the
two masses. In any case, the force of gravity is minuscule unless at least one of the
masses is extremely big.
The force of a large mass (such as the earth) on a mass many orders of magnitude
smaller (such as a person) is called weight, W . If the mass of the large object is M and
that of the relatively tiny one is m, then the weight of the small body is
W = G Mm
r2
= m
(
GM
r2
)
or
W = mg (1.4)
where
g = GM
r2
(1.5)
8 Chapter 1 Dynamics of point masses
g has units of acceleration (m/s2) and is called the acceleration of gravity. If planetary
gravity is the only force acting on a body, then the body is said to be in free fall. The
force of gravity draws a freely falling object towards the center of attraction (e.g.,
center of the earth) with an acceleration g . Under ordinary conditions, we sense our
own weight by feeling contact forces acting on us in opposition to the force of gravity.
In free fall there are, by definition, no contact forces, so there can be no sense of weight.
Even though the weight is not zero, a person in free fall experiences weightlessness,
or the absence of gravity.
Let us evaluate Equation 1.5 at the surface of the earth, whose radius according
to Table A.1 is 6378 km. Letting g0 represent the standard sea-level value of g , we get
g0 = GM
R2E
(1.6)
In SI units,
g0 = 9.807 m/s (1.7)
Substituting Equation 1.6 into Equation 1.5 and letting z represent the distance above
the earth’s surface, so that r = RE + z, we obtain
g = g0 R
2
E
(RE + z)2 =
g0
(1 + z/RE)2 (1.8)
Commercial airliners cruise at altitudes on the order of 10 kilometers (six miles). At
that height, Equation 1.8 reveals that g (and hence weight) is only three-tenths of a
percent less than its sea-level value. Thus, under ordinary conditions, we ignore the
variation of g with altitude. A plot of Equation 1.8 out to a height of 1000 km (the
upper limit of low-earth orbit operations) is shown in Figure 1.3. The variation of
g over that range is significant. Even so, at space station altitude (300 km), weight is
only about 10 percent less that it is on the earth’s surface. The astronauts experience
weightlessness, but they clearly are not weightless.
200 400 600 800
0.7
0.8
0.9
1.0
1000
z, km
g /
g 0
0
0
Figure 1.3 Variation of the acceleration of gravity with altitude.
louiscoo
铅笔
1.3 Mass, force and Newton’s law of gravitation 9
Example
1.3
Show that in the absence of an atmosphere, the shape of a low altitude ballistic
trajectory is a parabola. Assume the acceleration of gravity g is constant and neglect
the earth’s curvature.
x
y
(x0, y0)
υ0
P
g
g0
Figure 1.4 Flight of a low altitude projectile in free fall (no atmosphere).
Figure 1.4 shows a projectile launched at t = 0 with a speed v0 at a flight path angle
γ0 from the point with coordinates (x0, y0). Since the projectile is in free fall after
launch, its only acceleration is that of gravity in the negative y-direction:
ẍ = 0
ÿ = −g
Integrating with respect to time and applying the initial conditions leads to
x = x0 + (v0 cos γ0)t (a)
y = y0 + (v0 sin γ0)t − 1
2
gt2 (b)
Solving (a) for t and substituting the result into (b) yields
y = y0 + (x − x0) tan γ0 − 1
2
g
v0 cos γ0
(x − x0)2 (c)
This is the equation of a second-degree curve, a parabola, as sketched in Figure 1.4.
Example
1.4
An airplane flies a parabolic trajectory like that in Figure 1.4 so that the passengers
will experience free fall (weightlessness). What is the required variation of the flight
path angle γ with speed v? Ignore the curvature of the earth.
Figure 1.5 reveals that for a ‘flat’ earth, dγ = −dφ, i.e.,
γ̇ = −φ̇
10 Chapter 1 Dynamics of point masses
(Example 1.4
continued)
It follows from Equation 1.2 that
�γ̇ = −v (1.9)
The normal acceleration an is just the component of the gravitational acceleration g
in the direction of the unit principal normal to the curve (from P towards C). From
Figure 1.5, then,
an = g cos γ (a)
Substituting Equation 1.1 into (a) and solving for the radius of curvature yields
� = v
2
g cos γ
(b)
Combining Equations 1.9 and (b), we find the time rate of change of the flight path
angle,
γ̇ = −g cos γ
v
df
g
dg
C
r
y
x
g
P
g
Figure 1.5 Relationship between dγ and dφ for a ‘flat’ earth.
1.4 Newton’s law of motion
Force is not a primitive concept like mass because it is intimately connected with the
concepts of motion and inertia. In fact, the only way to alter the motion of a body is
to exert a force on it. The degree to which the motion is altered is a measure of the
force. This is quantified by Newton’s second law of motion. If the resultant or net
force on a body of mass m is Fnet,then
Fnet = ma (1.10)
1.4 Newton’s law of motion 11
x
y
z
r
Inertial frame
i
j
k
O
m
Fnet
a
v
ˆ
ˆ
ˆ
Figure 1.6 The absolute acceleration of a particle is in the direction of the net force.
In this equation, a is the absolute acceleration of the center of mass. The absolute
acceleration is measured in a frame of reference which itself has neither translational
nor rotational acceleration relative to the fixed stars. Such a reference is called an
absolute or inertial frame of reference.
Force, then, is related to the primitive concepts of mass, length and time by
Newton’s second law. The unit of force, appropriately, is the Newton, which is the
force required to impart an acceleration of 1 m/s2 to a mass of 1 kg. A mass of one
kilogram therefore weighs 9.81 Newtons at the earth’s surface. The kilogram is not a
unit of force.
Confusion can arise when mass is expressed in units of force, as frequently occurs
in US engineering practice. In common parlance either the pound or the ton (2000
pounds) is more likely to be used to express the mass. The pound of mass is officially
defined precisely in terms of the kilogram as shown in Table A.3. Since one pound of
mass weighs one pound of force where the standard sea-level acceleration of gravity
(g0 = 9.80665 m/s2) exists, we can use Newton’s second law to relate the pound of
force to the Newton:
1 lb (force) = 0.4536 kg × 9.807 m/s2
= 4.448 N
The slug is the quantity of matter accelerated at one foot per second2 by a force of
one pound. We can again use Newton’s second law to relate the slug to the kilogram.
Noting the relationship between feet and meters in Table A.3, we find
1 slug = 1 lb
1 ft/s2
= 4.448 N
0.3048 m/s2
= 14.59 kg · m/s
2
m/s2
= 14.59 kg
louiscoo
下划线
12 Chapter 1 Dynamics of point masses
Example
1.5
On a NASA mission the space shuttle Atlantis orbiter was reported to weigh 239 255 lb
just prior to lift-off. On orbit 18 at an altitude of about 350 km, the orbiter’s weight
was reported to be 236 900 lb. (a) What was the mass, in kilograms, of Atlantis on the
launch pad and in orbit? (b) If no mass were lost between launch and orbit 18, what
would have been the weight of Atlantis in pounds?
(a) The given data illustrates the common use of weight in pounds as a measure of
mass. The ‘weights’ given are actually the mass in pounds of mass. Therefore,
prior to launch
mlaunch pad = 239 255 lb (mass) × 0.4536 kg
1 lb (mass)
= 108 500 kg
In orbit,
morbit 18 = 236 900 lb (mass) × 0.4536 kg
1 lb (mass)
= 107 500 kg
The decrease in mass is the propellant expended by the orbital maneuvering and
reaction control rockets on the orbiter.
(b) Since the space shuttle launch pad at Kennedy Space Center is essentially at sea
level, the launch-pad weight of Atlantis in lb (force) is numerically equal to its
mass in lb (mass). With no change in mass, the force of gravity at 350 km would
be, according to Equation 1.8,
W = 239 255 lb (force) ×
(
1
1 + 3506378
)2
= 215 000 lb (force)
The integral of a force F over a time interval is called the impulse I of the force,
I =
∫ t2
t1
F dt (1.11)
From Equation 1.10 it is apparent that if the mass is constant, then
Inet =
∫ t2
t1
m
dv
dt
dt = mv2 − mv1 (1.12)
That is, the net impulse on a body yields a change m�v in its linear momentum,
so that
�v = Inet
m
(1.13)
If Fnet is constant, then Inet = Fnet�t , in which case Equation 1.13 becomes
�v = Fnet
m
�t (if Fnet is constant) (1.14)
1.4 Newton’s law of motion 13
Let us conclude this section by introducing the concept of angular momentum. The
moment of the net force about O in Figure 1.6 is
MOnet = r × Fnet
Substituting Equation 1.10 yields
MOnet = r × ma = r × m
dv
dt
(1.15)
But, keeping in mind that the mass is constant,
r × m dv
dt
= d
dt
(r × mv) −
(
dr
dt
× mv
)
= d
dt
(r × mv) − (v × mv)
Since v × mv = m(v × v) = 0, it follows that Equation 1.15 can be written
MOnet =
dHO
dt
(1.16)
where HO is the angular momentum about O,
HO = r × mv (1.17)
Thus, just as the net force on a particle changes its linear momentum mv, the moment
of that force about a fixed point changes the moment of its linear momentum about
that point. Integrating Equation 1.16 with respect to time yields∫ t2
t1
MOnet dt = HO2 − HO1 (1.18)
The integral on the left is the net angular impulse. This angular impulse–momentum
equation is the rotational analog of the linear impulse–momentum relation given
above in Equation 1.12.
Example
1.6
A particle of mass m is attached to point O by an inextensible string of length l.
Initially the string is slack when m is moving to the left with a speed vo in the position
shown. Calculate the speed of m just after the string becomes taut. Also, compute the
x
y
d
υ0
υ
l
m
O
c
Figure 1.7 Particle attached to O by an inextensible string.
14 Chapter 1 Dynamics of point masses
(Example 1.6
continued)
average force in the string over the small time interval �t required to change the
direction of the particle’s motion.
Initially, the position and velocity of the particle are
r1 = c î + dĵ v1 = −v0 î
The angular momentum is
H1 = r1 × mv1 =
∣∣∣∣∣∣
î ĵ k̂
c d 0
−mv0 0 0
∣∣∣∣∣∣ = mv0k̂ (a)
Just after the string becomes taut
r2 = −
√
l2 − d2 î + dĵ v2 = vx î + vy ĵ (b)
and the angular momentum is
H2 = r2 × mv2 =
∣∣∣∣∣∣
î ĵ k̂
−√l2 − d2 d 0
vx vy 0
∣∣∣∣∣∣ =
(
−mvxd − mvy
√
l2 − d2
)
k̂ (c)
Initially the force exerted on m by the slack string is zero. When the string becomes
taut, the force exerted on m passes through O. Therefore, the moment of the net force
on m about O remains zero. According to Equation 1.18,
H2 = H1
Substituting (a) and (c) yields
vxd +
√
l2 − d2vy = −v0d (d)
The string is inextensible, so the component of the velocity of m along the string must
be zero:
v2 · r2 = 0
Substituting v2 and r2 from (b) and solving for vy we get
vy = vx
√
l2
d2
− 1 (e)
Solving (d) and (e) for vx and vy leads to
vx = −d
2
l2
v0 vy = −
√
1 − d
2
l2
d
l
v0 (f)
Thus, the speed, v =
√
v2x + v2y , after the string becomes taut is
v = d
l
v0
1.5 Time derivatives of moving vectors 15
From Equation 1.12, the impulse on m during the time it takes the string to become
taut is
I = m(v2 − v1) = m
[(
−d
2
l2
v0 î −
√
1 − d
2
l2
d
l
v0 ĵ
)
− (−v0 î)
]
=
(
1 − d
2
l2
)
mv0 î −
√
1 − d
2
l2
d
l
mv0 ĵ
The magnitude of this impulse, which is directed along the string, is
I =
√
1 − d
2
l2
mv0
Hence, the average force in the string during the small time interval �t required to
change the direction of the velocity vector turns out to be
Favg = I
�t
=
√
1 − d
2
l2
mv0
�t
1.5 Time derivatives of moving vectors
Figure 1.8(a) shows a vector A inscribed in a rigid body B that is in motion relative
to an inertial frame of reference (a rigid, cartesian coordinate system which is fixed
relative to the fixed stars). The magnitude of A is fixed. The body B is shown at two
times, separated by the differential time interval dt . At time t + dt the orientation of
ω
f
dθ
A
dA
A � dA
(a) (b)
A(t � dt)
X
Y
Z
t t � dt
A(t)
Rigid body B 
Inertial frame
Instantaneous axis of rotation
Figure 1.8 Displacement of a rigid body.
16 Chapter 1 Dynamics of point masses
vector A differs slightly from that at time t , but its magnitude is the same. According
to one of the many theorems of the prolific eighteenth century Swiss mathematician
Leonhard Euler (1707–1783), there is a unique axis of rotation about which B and,
therefore, A rotates during the differential time interval. If we shift the two vectors
A(t) and A(t + dt) to the same point on the axis of rotation, so that they are tail-to-tail
as shown in Figure 1.8(b), we can assess the difference dA between them caused by
the infinitesimal rotation. Remember that shifting a vector to a parallel line does not
change the vector. The rotation of the body B is measured in the plane perpendicular
to the instantaneous axis of rotation. The amount of rotation is the angle dθ through
whicha line element normal to the rotation axis turns in the time interval dt . In
Figure 1.8(b) that line element is the component of A normal to the axis of rotation.
We can express the difference dA between A(t) and A(t + dt) as
dA =
magnitude of dA︷ ︸︸ ︷
[(‖A‖ · sin φ)dθ] n̂ (1.19)
where n̂ is the unit normal to the plane defined by A and the axis of rotation, and
it points in the direction of the rotation. The angle φ is the inclination of A to the
rotation axis. By definition,
dθ = ‖ω‖dt (1.20)
where ω is the angular velocity vector, which points along the instantaneous axis of
rotation and its direction is given by the right-hand rule. That is, wrapping the right
hand around the axis of rotation, with the fingers pointing in the direction of dθ ,
results in the thumb’s defining the direction of ω. This is evident in Figure 1.8(b). It
should be pointed out that the time derivative of ω is the angular acceleration, usually
given the symbol α. Thus,
α = dω
dt
(1.21)
Substituting Equation 1.20 into Equation 1.19, we get
dA = ‖A‖ · sin φ‖ω‖dt · n̂ = (‖ω‖ · ‖A‖ · sin φ) n̂ dt (1.22)
By definition of the cross product, ω × A is the product of the magnitude of ω, the
magnitude of A, the sine of the angle between ω and A and the unit vector normal to
the plane of ω and A, in the rotation direction. That is,
ω × A = ‖ω‖ · ‖A‖ · sin φ · n̂ (1.23)
Substituting Equation 1.23 into Equation 1.22 yields
dA = ω × Adt
Dividing through by dt , we finally obtain
dA
dt
= ω × A (1.24)
Equation 1.24 is a formula we can use to compute the time derivative of any vector
of constant magnitude.
1.5 Time derivatives of moving vectors 17
Example
1.7
Calculate the second time derivative of a vector A of constant magnitude, expressing
the result in terms of ω and its derivatives and A.
Differentiating Equation 1.24 with respect to time, we get
d2A
dt2
= d
dt
dA
dt
= d
dt
(ω × A) = dω
dt
× A + ω × dA
dt
Using Equations 1.21 and 1.24, this can be written
d2A
dt2
= α × A + ω × (ω × A) (1.25)
Example
1.8
Calculate the third derivative of a vector A of constant magnitude, expressing the
result in terms of ω and its derivatives and A.
d3A
dt3
= d
dt
d2A
dt2
= d
dt
[α × A + ω × (ω × A)]
= d
dt
(α × A) + d
dt
[ω × (ω × A)]
=
(
dα
dt
× A + α × dA
dt
)
+
[
dω
dt
× (ω × A) + ω × d
dt
(ω × A)
]
=
[
dα
dt
× A + α × (ω × A)
]
+
[
α × (ω × A) + ω ×
(
dω
dt
× A + ω × dA
dt
)]
=
[
dα
dt
× A + α × (ω × A)
]
+ {α × (ω × A) + ω × [α × A + ω × (ω × A)]}
= dα
dt
× A + α × (ω × A) + α × (ω × A) + ω × (α × A) + ω × [ω × (ω × A)]
= dα
dt
× A + 2α × (ω × A) + ω × (α × A) + ω × [ω × (ω × A)]
d3A
dt3
= dα
dt
× A + 2α × (ω × A) + ω × [α × A + ω × (ω × A)]
Let XYZ be a rigid inertial frame of reference and xyz a rigid moving frame of
reference, as shown in Figure 1.9. The moving frame can be moving (translating and
rotating) freely of its own accord, or it can be imagined to be attached to a physical
object, such as a car, an airplane or a spacecraft. Kinematic quantities measured
relative to the fixed inertial frame will be called absolute (e.g., absolute acceleration),
and those measured relative to the moving system will be called relative (e.g., relative
acceleration). The unit vectors along the inertial XYZ system are Î, Ĵ and K̂, whereas
those of the moving xyz system are î, ĵ and k̂. The motion of the moving frame is
arbitrary, and its absolute angular velocity is �. If, however, the moving frame is
rigidly attached to an object, so that it not only translates but rotates with it, then the
18 Chapter 1 Dynamics of point masses
X
Y
Z
x
y
z
Inertial frame
Moving frame
O
Q
Qy
Qx
Qz
J
I
K
i
j 
k 
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
Figure 1.9 Fixed (inertial) and moving rigid frames of reference.
frame is called a body frame and the axes are referred to as body axes. A body frame
clearly has the same angular velocity as the body to which it is bound.
Let Q be any time-dependent vector. Resolved into components along the inertial
frame of reference, it is expressed analytically as
Q = QX Î + QY Ĵ + QZ K̂
where QX , QY and QZ are functions of time. Since Î, Ĵ and K̂ are fixed, the time
derivative of Q is simply given by
dQ
dt
= dQX
dt
Î + dQY
dt
Ĵ + dQZ
dt
K̂
dQX/dt , dQY /dt and dQZ/dt are the components of the absolute time derivative of Q.
Q may also be resolved into components along the moving xyz frame, so that, at
any instant,
Q = Qx î + Qy ĵ + Qz k̂ (1.26)
Using this expression to calculate the time derivative of Q yields
dQ
dt
= dQx
dt
î + dQy
dt
ĵ + dQz
dt
k̂ + Qx dî
dt
+ Qy dĵ
dt
+ Qz dk̂
dt
(1.27)
The unit vectors î, ĵ and k̂ are not fixed in space, but are continuously changing
direction; therefore, their time derivatives are not zero. They obviously have a constant
1.5 Time derivatives of moving vectors 19
magnitude (unity) and, being attached to the xyz frame, they all have the angular
velocity �. It follows from Equation 1.24 that
dî
dt
= � × î dĵ
dt
= � × ĵ dk̂
dt
= � × k̂
Substituting these on the right-hand side of Equation 1.27 yields
dQ
dt
= dQx
dt
î + dQy
dt
ĵ + dQz
dt
k̂ + Qx(� × î) + Qy(� × ĵ) + Qz(� × k̂)
= dQx
dt
î + dQy
dt
ĵ + dQz
dt
k̂ + (� × Qx î) + (� × Qy ĵ) + (� × Qz k̂)
= dQx
dt
î + dQy
dt
ĵ + dQz
dt
k̂ + � × (Qx î + Qy ĵ + Qz k̂)
In view of Equation 1.26, this can be written
dQ
dt
= dQ
dt
)
rel
+ � × Q (1.28)
where
dQ
dt
)
rel
= dQx
dt
î + dQy
dt
ĵ + dQz
dt
k̂ (1.29)
dQ/dt)rel is the time derivative of Q relative to the moving frame. Equation 1.28 shows
how the absolute time derivative is obtained from the relative time derivative. Clearly,
dQ/dt = dQ/dt)rel only when the moving frame is in pure translation (� = 0).
Equation 1.28 can be used recursively to compute higher order time derivatives.
Thus, differentiating Equation 1.28 with respect to t , we get
d2Q
dt2
= d
dt
dQ
dt
)
rel
+ d�
dt
× Q + � × dQ
dt
Using Equation 1.28 in the last term yields
d2Q
dt2
= d
dt
dQ
dt
)
rel
+ d�
dt
× Q + � ×
[
dQ
dt
)
rel
+ � × Q
]
(1.30)
Equation 1.28 also implies that
d
dt
dQ
dt
)
rel
= d
2Q
dt2
)
rel
+ � × dQ
dt
)
rel
(1.31)
where
d2Q
dt2
)
rel
= d
2Qx
dt2
î + d
2Qy
dt2
ĵ + d
2Qz
dt2
k̂
Substituting Equation 1.31 into Equation 1.30 yields
d2Q
dt2
=
[
d2Q
dt2
)
rel
+ � × dQ
dt
)
rel
]
+ d�
dt
× Q + � ×
[
dQ
dt
)
rel
+ � × Q
]
(1.32)
20 Chapter 1 Dynamics of point masses
Collecting terms, this becomes
d2Q
dt2
= d
2Q
dt2
)
rel
+ �̇ × Q + � × (� × Q) + 2� × dQ
dt
)
rel
where �̇ ≡ d�/dt is the absolute angular acceleration of the xyz frame.
Formulas for higher order time derivatives are found in a similar fashion.
1.6 Relative motion
Let P be a particle in arbitrary motion. The absolute position vector of P is r and the
position of P relative to the moving frame is rrel. If rO is the absolute position of the
origin of the moving frame, then it is clear from Figure 1.10 that
r = rO + rrel (1.33)
Since rrel is measured in the moving frame,
rrel = xî + y ĵ + zk̂ (1.34)
where x, y and z are the coordinates of P relative to the moving reference.
The absolute velocity v of P is dr/dt , so that from Equation 1.33 we have
v = vO + drrel
dt
(1.35)
where vO = drO/dt is the (absolute) velocity of the origin of the xyz frame. From
Equation 1.28, we can write
drrel
dt
= vrel + � × rrel (1.36)
X
Y
Z
x
y
z
r
Inertial frame
(non-rotating, non-accelerating)
Moving frame
O
P
r0
rrel
J
I
K
j 
k 
i
ˆ
ˆ
ˆ
ˆ
ˆ
Figure 1.10 Absolute and relative position vectors.
1.6 Relative motion 21
where vrel is the velocity of P relative to the xyz frame:
vrel = drrel
dt
)
rel
= dx
dt
î + dy
dt
ĵ + dz
dt
k̂ (1.37)
Substituting Equation 1.36 into Equation 1.35 yields
v = vO + � × rrel + vrel (1.38)
The absolute acceleration a of P is dv/dt , so that from Equation 1.35 we have
a = aO + d
2rrel
dt2
(1.39)
where aO = dvO/dt is the absolute acceleration of the origin of the xyz frame. We
evaluatethe second term on the right using Equation 1.32:
d2rrel
dt2
= d
2rrel
dt2
)
rel
+ �̇ × rrel + � × (� × rrel) + 2� × drrel
dt
)
rel
(1.40)
Since vrel = drrel/dt)rel and arel = d2rrel/dt2)rel, this can be written
d2rrel
dt2
= arel + �̇ × rrel + � × (� × rrel) + 2� × vrel (1.41)
Upon substituting this result into Equation 1.39, we find
a = aO + �̇ × rrel + � × (� × rrel) + 2� × vrel + arel (1.42)
The cross product 2�×vrel is called the Coriolis acceleration after Gustave Gaspard de
Coriolis (1792–1843), the French mathematician who introduced this term (Coriolis,
1835). For obvious reasons, Equation 1.42 is sometimes referred to as the five-term
acceleration formula.
Example
1.9
At a given instant, the absolute position, velocity and acceleration of the origin O of
a moving frame are
rO = 100Î + 200Ĵ + 300K̂ (m)
vO = −50Î + 30Ĵ − 10K̂ (m/s) (given) (a)
aO = −15Î + 40Ĵ + 25K̂ (m/s2)
The angular velocity and acceleration of the moving frame are
� = 1.0Î − 0.4Ĵ + 0.6K̂ (rad/s)
(given) (b)
�̇ = −1.0Î + 0.3Ĵ − 0.4K̂ (rad/s2)
The unit vectors of the moving frame are
î = 0.5571Î + 0.7428Ĵ + 0.3714K̂
ĵ = −0.06331Î + 0.4839Ĵ − 0.8728K̂ (given) (c)
k̂ = −0.8280Î + 0.4627Ĵ + 0.3166K̂
22 Chapter 1 Dynamics of point masses
(Example 1.9
continued)
The absolute position, velocity and acceleration of P are
r = 300Î − 100Ĵ + 150K̂ (m)
v = 70Î + 25Ĵ − 20K̂ (m/s) (given) (d)
a = 7.5Î − 8.5Ĵ + 6.0K̂ (m/s2)
Find the velocity vrel and acceleration arel of P relative to the moving frame.
First use Equations (c) to solve for Î, Ĵ and K̂ in terms of î, ĵ and k̂ (three equations
in three unknowns):
Î = 0.5571î − 0.06331ĵ − 0.8280k̂
Ĵ = 0.7428î + 0.4839ĵ + 0.4627k̂ (e)
K̂ = 0.3714î − 0.8728ĵ + 0.3166k̂
The relative position vector is
rrel = r − rO = (300Î − 100Ĵ + 150K̂) − (100Î + 200Ĵ + 300K̂)
= 200Î − 300Ĵ − 150K̂ (m) (f)
From Equation 1.38, the relative velocity vector is
vrel = v − vO − � × rrel
= (70Î + 25Ĵ − 20K̂) − (−50Î + 30Ĵ − 10K̂) −
∣∣∣∣∣∣
Î Ĵ K̂
1.0 −0.4 0.6
200 −300 −150
∣∣∣∣∣∣
= (70Î + 25Ĵ − 20K̂) − (−50Î + 30Ĵ − 10K̂) − (240Î + 270Ĵ − 220K̂)
or
vrel = −120Î − 275Ĵ + 210K̂ (m/s) (g)
To obtain the components of the relative velocity along the axes of the moving frame,
substitute Equations (e) into Equation (g).
vrel = −120(0.5571i − 0.06331j − 0.8280k)
−275(0.7428i + 0.4839j + 0.4627k) + 210(0.3714i − 0.8728j + 0.3166k)
so that
vrel = −193.1î − 308.8ĵ + 38.60k̂ (m/s) (h)
Alternatively,
vrel = 366.2ûv (m/s), where ûv = −0.5272î − 0.8432ĵ + 0.1005k̂ (i)
1.6 Relative motion 23
To find the relative acceleration, we use the five-term acceleration formula,
Equation 1.42:
arel = a − aO − �̇ × rrel − � × (� × rrel) − 2(� × vrel)
= a − aO −
∣∣∣∣∣∣
Î Ĵ K̂
−1.0 0.3 −0.4
200 −300 −150
∣∣∣∣∣∣ − �
×
∣∣∣∣∣∣
Î Ĵ K̂
1.0 −0.4 0.6
200 −300 −150
∣∣∣∣∣∣ − 2
∣∣∣∣∣∣
Î Ĵ K̂
1.0 −0.4 0.6
−120 −275 210
∣∣∣∣∣∣
= a − aO − (−165Î − 230Ĵ + 240K̂) −
∣∣∣∣∣∣
Î Ĵ K̂
1.0 −0.4 0.6
240 270 −220
∣∣∣∣∣∣
− (162Î − 564Ĵ − 646K̂)
= (7.5Î − 8.5Ĵ + 6K̂) − (−15Î + 40Ĵ + 25K̂)
− (−165Î − 230Ĵ + 240K̂) − (−74Î + 364Ĵ + 366K̂)
− (162Î − 564Ĵ − 646K̂)
arel = 99.5Î + 381.5Ĵ + 21.0K̂ (m/s2) (j)
The components of the relative acceleration along the axes of the moving frame are
found by substituting Equations (e) into Equation (j):
arel = 99.5(0.5571î − 0.06331ĵ − 0.8280k̂)
+ 381.5(0.7428î + 0.4839ĵ + 0.4627k̂) + 21.0(0.3714î − 0.8728ĵ + 0.3166k̂)
arel = 346.6î + 160.0ĵ + 100.8k̂ (m/s2) (k)
or
arel = 394.8ûa (m/s2), where ûa = 0.8778î + 0.4052ĵ + 0.2553k̂ (l)
Figure 1.11 shows the non-rotating inertial frame of reference XYZ with its origin
at the center C of the earth, which we shall assume to be a sphere. That assumption
will be relaxed in Chapter 5. Embedded in the earth and rotating with it is the
orthogonal x′y′z′ frame, also centered at C, with the z′ axis parallel to Z , the earth’s
axis of rotation. The x′ axis intersects the equator at the prime meridian (zero degrees
longitude), which passes through Greenwich in London, England. The angle between
X and x′ is θg , and the rate of increase of θg is just the angular velocity � of the earth.
P is a particle (e.g., an airplane, spacecraft, etc.), which is moving in an arbitrary
fashion above the surface of the earth. rrel is the position vector of P relative to C in
the rotating x′y′z′ system. At a given instant, P is directly over point O, which lies on
24 Chapter 1 Dynamics of point masses
X
Y
Λ
(East longitude)
Ω
x (East)
y (North) z (Zenith)
Earth
C
 x ′
 y ′
Z, z ′
 P
RE
O
θg
Greenwich meridian
Equator
r rel
j k 
i
φ (North latitude)
ˆ
ˆ
ˆ
Figure 1.11 Earth-centered inertial frame (XYZ); earth-centered non-inertial x′y′z′ frame embedded in
and rotating with the earth; and a non-inertial, topocentric-horizon frame xyz attached to a
point O on the earth’s surface.
the earth’s surface at longitude � and latitude φ. Point O coincides instantaneously
with the origin of what is known as a topocentric-horizon coordinate system xyz.
For our purposes x and y are measured positive eastward and northward along the
local latitude and meridian, respectively, through O. The tangent plane to the earth’s
surface at O is the local horizon. The z axis is the local vertical (straight up) and
it is directed radially outward from the center of the earth. The unit vectors of the
xyz frame are îĵk̂, as indicated in Figure 1.11. Keep in mind that O remains directly
below P, so that as P moves, so do the xyz axes. Thus, the îĵk̂ triad, which are the
unit vectors of a spherical coordinate system, vary in direction as P changes location,
thereby accounting for the curvature of the earth.
Let us find the absolute velocity and acceleration of P. It is convenient first to
obtain the velocity and acceleration of P relative to the non-rotating earth, and then
use Equations 1.38 and 1.42 to calculate their inertial values.
The relative position vector can be written
rrel = (RE + z)k̂ (1.43)
1.6 Relative motion 25
where RE is the radius of the earth and z is the height of P above the earth (i.e., its
altitude). The time derivative of rrel is the velocity vrel relative to the non-rotating
earth,
vrel = drrel
dt
= żk̂ + (RE + z) dk̂
dt
(1.44)
To calculate dk̂/dt , we must use Equation 1.24. The angular velocity ω of the xyz
frame relative to the non-rotating earth is found in terms of the rates of change of
latitude φ and longitude �,
ω = −φ̇ î + �̇ cos φ ĵ + �̇ sin φk̂ (1.45)
Thus,
dk̂
dt
= ω × k̂ = �̇ cos φ î + φ̇ ĵ (1.46)
Let us also record the following for future use:
dĵ
dt
= ω × ĵ = −�̇ sin φ ĵ − φ̇k̂ (1.47)
dî
dt
= ω × î = �̇ sin φ ĵ − �̇ cos φk̂ (1.48)
Substituting Equation 1.46 into Equation 1.44 yields
vrel = ẋî + ẏ ĵ + żk̂ (1.49a)
where
ẋ = (RE + z)�̇ cos φ ẏ = (RE + z)φ̇ (1.49b)
It is convenient to use these results to express the rates of change of latitude and
longitude in terms of the components of relative velocity over the earth’s surface,
φ̇ = ẏ
RE + z �̇ =
ẋ
(RE + z) cos φ (1.50)
The time derivatives of these two expressions are
φ̈ = (RE + z)ÿ − ẏż
(RE + z)2 �̈ =
(RE + z)ẍ cos φ − (ż cos φ − ẏ sin φ)ẋ
(RE + z)2 cos2 φ (1.51)
The acceleration of P relative to the non-rotating earth is found by taking the time
derivative of vrel. From Equation 1.49 we thereby obtain
arel = ẍî + ÿ ĵ + z̈k̂ + ẋ dî
dt
+ ẏ dĵ
dt
+ ż dk̂
dt
= [ż�̇ cos φ + (RE + z)�̈ cos φ − (RE + z)φ̇�̇ sin φ]î
+ [żφ̇ + (RE + z)φ̈]ĵ + z̈k̂ + (RE + z)�̇ cos φ(ω × î)
+ (RE + z)φ̇(ω × ĵ) + ż(ω × k̂)
26 Chapter 1 Dynamics of point masses
Substituting Equations 1.46 through 1.48 together with 1.50 and 1.51 into this
expression yields, upon simplification,
arel =
[̈
x + ẋ(ż − ẏ tan φ)
RE + z
]
î +
(
ÿ + ẏż + ẋ
2 tan φ
RE + z
)
ĵ +
(̈
z − ẋ
2 + ẏ2
RE + z
)
k̂ (1.52)
Observe that the curvature of the earth’s surface is neglected by letting RE + z become
infinitely large,in which case
arel)neglecting earth′s curvature = ẍî + ÿ ĵ + z̈k̂
That is, for a ‘flat earth’, the components of the relative acceleration vector are just the
derivatives of the components of the relative velocity vector.
For the absolute velocity we have, according to Equation 1.38,
v = vC + � × rrel + vrel (1.53)
From Figure 1.11 it can be seen that K̂ = cos φ ĵ + sin φk̂, which means the angular
velocity of the earth is
� = �K̂ = � cos φ ĵ + � sin φk̂ (1.54)
Substituting this, together with Equations 1.43 and 1.49a and the fact that vC = 0,
into Equation 1.53 yields
v = [ẋ + �(RE + z) cos φ]î + ẏ ĵ + żk̂ (1.55)
From Equation 1.42 the absolute acceleration of P is
a = aC + �̇ × rrel + � × (� × rrel) + 2� × vrel + arel
Since aC = �̇ = 0, we find, upon substituting Equations 1.43, 1.49a, 1.52 and 1.54, that
a =
[
ẍ + ẋ(ż − ẏ tan φ)
RE + z + 2�(ż cos φ − ẏ sin φ)
]
î
+
{
ÿ + ẏż + ẋ
2 tan φ
RE + z + � sin φ[�(RE + z) cos φ + 2ẋ]
}
ĵ
+
{
z̈ − ẋ
2 + ẏ2
RE + z − � cos φ[�(RE + z) cos φ + 2ẋ]
}
k̂ (1.56)
Some special cases of Equations 1.55 and 1.56 follow.
1.6 Relative motion 27
Straight and level, unaccelerated flight: ż = z̈ = ẍ = ÿ = 0
v = [ẋ + �(RE + z) cos φ]î + ẏ ĵ (1.57a)
a = −
[
ẋẏ tan φ
RE + z + 2�ẏ sin φ
]
î
+
{
ẋ2 tan φ
RE + z + � sin φ[�(RE + z) cos φ + 2ẋ]
}
ĵ
−
{
ẋ2 + ẏ2
RE + z + � cos φ[�(RE + z) cos φ + 2ẋ]
}
k̂ (1.57b)
Flight due north (y) at constant speed and altitude: ż = z̈ = ẋ = ẍ = ÿ = 0
v = �(RE + z) cos φ î + ẏ ĵ (1.58a)
a = −2�ẏ sin φ î + �2(RE + z) sin φ cos φ ĵ
−
[
ẏ2
RE + z + �
2(RE + z) cos2 φ
]
k̂ (1.58b)
Flight due east (x) at constant speed and altitude: ż = z̈ = ẍ = ẏ = ÿ = 0
v = [ẋ + �(RE + z) cos φ]î (1.59a)
a =
{
ẋ2 tan φ
RE + z + � sin φ [�(RE + z) cos φ + 2ẋ]
}
ĵ
−
{
ẋ2
RE + z + � cos φ [�(RE + z) cos φ + 2ẋ]
}
k̂ (1.59b)
Flight straight up (z): ẋ = ẍ = ẏ = ÿ = 0
v = �(RE + z) cos φ î + żk̂ (1.60a)
a = 2�(ż cos φ)î + �2(RE + z) sin φ cos φ ĵ
+ [z̈ − �2(RE + z) cos2 φ] k̂ (1.60b)
Stationary: ẋ = ẍ = ẏ = ÿ = ż = z̈ = 0
v = �(RE + z) cos φ î (1.61a)
a = �2(RE + z) sin φ cos φ ĵ − �2(RE + z) cos2 φk̂ (1.61b)
Example
1.10
An airplane of mass 70 000 kg is traveling due north at latitude 30◦ north, at an altitude
of 10 km (32 800 ft) with a speed of 300 m/s (671 mph). Calculate (a) the components
of the absolute velocity and acceleration along the axes of the topocentric-horizon
reference frame, and (b) the net force on the airplane.
28 Chapter 1 Dynamics of point masses
(Example 1.10
continued)
(a) First, using the sidereal rotation period of the earth in Table A.1, we note that the
earth’s angular velocity is
� = 2πrad
sidereal day
= 2πrad
23.93 hr
= 2πrad
86 160 s
= 7.292 × 10−5rad/s (a)
From Equation 1.58a, the absolute velocity is
v = �(RE + z) cos φ î + ẏ ĵ =
[
(7.292 × 10−5) · (6378 + 10) · 103 cos 30◦] î + 300ĵ
or
v = 403.4î + 300ĵ (m/s)
The 403.4 m/s (901 mph) component of velocity to the east (x direction) is due
entirely to the earth’s rotation.
From Equation 1.58b2, the absolute acceleration is
a = −2�ẏ sin φ î + �2(RE + z) sin φ cos φ ĵ −
[
ẏ2
RE + z + �
2(RE + z) cos2 φ
]
k̂
= −2(7.292 × 10−5) · 300 · sin 30◦ î
+ (7.292 × 10−5)2 · (6378 + 10) · 103 · sin 30◦ · cos 30◦ ĵ
−
[
3002
(6378 + 10) · 103 + (7.292 × 10
−5)2 · (6378 + 10) · 103 · cos2 30◦
]
k̂
or
a = −0.02187î + 0.01471ĵ − 0.03956k̂ (m/s2) (a)
The westward acceleration of 0.02187 m/s2 is the Coriolis acceleration.
(b) Since the acceleration in part (a) is the absolute acceleration, we can use it in
Newton’s law to calculate the net force on the airplane,
Fnet = ma = 70 000(−0.02187î + 0.01471ĵ − 0.03956k̂)
= −1531î + 1029ĵ − 2769k̂ (N)
Figure 1.12 shows the components of this relatively small force. The forward and
downward forces are in the directions of the airplane’s centripetal acceleration,
caused by the earth’s rotation and, in the case of the downward force, by the
earth’s curvature as well. The westward force is in the direction of the Coriolis
acceleration, which is due to the combined effects of the earth’s rotation and the
motion of the airplane. These net external forces must exist if the airplane is to
fly the prescribed path.
In the vertical direction, the net force is that of the upward lift L of the wings
plus the downward weight W of the aircraft, so that
Fnet)z = L − W = −2769 ⇒ L = W−2769 (N)
Problems 29
2769 N
(622 lb)
1531 N
(344 lb)
1029 N
(231 lb)
x
East
z
Up
y
North
Figure 1.12 Components of the net force on the airplane.
Thus, the effect of the earth’s rotation and curvature is to apparently produce an
outward centrifugal force, reducing the weight of the airplane a bit, in this case
by about 0.4 percent. The fictitious centrifugal force also increases the apparent
drag in the flight direction by 1029 N. That is, in the flight direction
Fnet)y = T − D = −2769 N
where T is the thrust and D is the drag. Hence
T = D + 1029 (N)
The 1531 N force to the left, produced by crabbing the airplane very slightly in
that direction, is required to balance the fictitious Coriolis force which would
otherwise cause the airplane to deviate to the right of its flight path.
Problems
1.1 Given the three vectors A = Ax î+Ay ĵ+Az k̂ B = Bx î+By ĵ+Bz k̂ C = Cx î+Cy ĵ+Cz k̂
show, analytically, that
(a) A · A = A2
(b) A · (B × C) = (A × B) · C (interchangeability of the ‘dot’ and ‘cross’)
(c) A × (B × C) = B(A · C) − C(A · B) (the bac – cab rule)
30 Chapter 1 Dynamics of point masses
(Simply compute the expressions on each side of the = signs and demonstrate conclusively
that they are the same. Do not substitute numbers to‘prove’your point. Use the fact that the
cartesian coordinate unit vectors î, ĵ and k̂ form a right-handed orthogonal triad, so that
î · ĵ = î · k̂ = ĵ · k̂ = 0 î · î = ĵ · ĵ = k̂ · k̂ = 1
î × ĵ = k̂ ĵ × k̂ = î k̂ × î = ĵ (î × k̂ = −ĵ ĵ × î = −k̂ k̂ × ĵ = −î)
Also,
î × î = ĵ × ĵ = k̂ × k̂ = 0
1.2 Use just the vector identities in parts (a) and (b) of Exercise 1.1 to show that
(A × B) · (C × D) = (A · C)(B · D) − (A · D)(B · C)
1.3 The absolute position, velocity and acceleration of O are
rO = 300Î + 200Ĵ + 100K̂ (m)
vO = −10Î + 30Ĵ − 50K̂ (m/s)
aO = 25Î + 40Ĵ − 15K̂ (m/s2)
The angular velocity and acceleration of the moving frame are
� = 0.6Î − 0.4Ĵ + 1.0K̂ (rad/s)
�̇ = −0.4Î + 0.3Ĵ − 1.0K̂ (rad/s2)
The unit vectors of the moving frame are
î = 0.57735Î + 0.57735Ĵ + 0.57735K̂
ĵ = −0.74296Î + 0.66475Ĵ + 0.078206K̂
k̂ = −0.33864Î − 0.47410Ĵ + 0.81274K̂
The absolute position of P is
r = 150Î − 200Ĵ + 300K̂ (m)
The velocity and acceleration of P relative to the moving frame are
vrel = −20î + 25ĵ + 70k̂ (m/s) arel = 7.5î − 8.5ĵ + 6.0k̂ (m/s2)
Calculate the absolute velocity vP and acceleration aP of P.
{Ans.: vP = 478.7ûv (m/s), ûv = 0.5352Î − 0.5601Ĵ − 0.6324K̂;
aP = 616.3ûa (m/s2), ûa = 0.1655Î + 0.9759Ĵ + 0.1424K̂}
1.4 F is a force vector of fixed magnitude embedded on a rigid body in plane motion (in the
xy plane). At a given instant, ω= 3k̂ rad/s, ω̇= −2k̂ rad/s2, ω̈= 0 and F = 10î N. At that
instant, calculate
...
F .
{Ans.:
...
F = 180î − 270ĵ N/s3}
Problems 31
X
Y
Z
x
y
z
r
Inertial frame
Moving frame
O
P
r0
rrel
Figure P.1.3
υ
θ
r
x
y
h
X
Y
A
Figure P.1.5
1.5 An airplane in level flight at an altitude h and a uniform speed v passes directly over
a radar tracking station A. Calculate the angular velocity θ̇ and angular acceleration of
the radar antenna θ̈ as well as the rate ṙ at which the airplane is moving away from the
antenna. Use the equations of this chapter (rather than polar coordinates, which you can
use to check your work). Attach the inertial frame of reference to the ground and assume
a non-rotating earth. Attach the moving frame to the antenna, with the x axis pointing
always from the antenna towards the airplane.
{Ans.: (a) θ̇ = v cos2 θ /h; (b)θ̈ = −2v2 cos3 θ sin θ /h2; (c) vrel = v sin θ}
32 Chapter 1 Dynamics of point masses
1.6 At 30◦ north latitude, a 1000 kg (2205 lb) car travels due north at a constant speed of
100 km/hr (62 mph) on a level road at sea level. Taking into account the earth’s rotation,
calculate the lateral (sideways) force of the road on the car, and the normal force of the
road on the car.
{Ans.: Flateral = 2.026 N, to the left (west); N = 9784 N}
1.7 At 29◦ north latitude, what is the deviation d from the vertical of a plumb bob at the end
of a 30 m string, due to the earth’s rotation?
{Ans.: 44.1 mm to the south}
θ
g
y
z
North d
L � 30 m
Figure P.1.7
2C h a p t e r
The two-body
problem
Chapter outline
2.1 Introduction 33
2.2 Equations of motion in an inertial frame 34
2.3 Equations of relative motion 37
2.4 Angular momentum and the orbit formulas 42
2.5 The energy law 50
2.6 Circular orbits (e = 0) 51
2.7 Elliptical Orbits (0 < e < 1) 55
2.8 Parabolic trajectories (e = 1) 65
2.9 Hyperbolic trajectories (e > 1) 69
2.10 Perifocal frame 76
2.11 The Lagrange coefficients 78
2.12 Restricted three-body problem 89
2.12.1 Lagrange points 92
2.12.2 Jacobi constant 96
Problems 101
2.1 Introduction
This chapter presents the vector-based approach to the classical problem of deter-mining the motion of two bodies due solely to their own mutual gravitational
attraction. We show that the path of one of the masses relative to the other is a
conic section (circle, ellipse, parabola or hyperbola) whose shape is determined by
the eccentricity. Several fundamental properties of the different types of orbits are
33
34 Chapter 2 The two-body problem
developed with the aid of the laws of conservation of angular momentum and energy.
These properties include the period of elliptical orbits, the escape velocity associated
with parabolic paths and the characteristic energy of hyperbolic trajectories. Follow-
ing the presentation of the four types of orbits, the perifocal frame is introduced. This
frame of reference is used to describe orbits in three dimensions, which is the subject
of Chapter 4.
In this chapter the perifocal frame provides the backdrop for developing the
Lagrange f and g coefficients. By means of the Lagrange f and g coefficients, the posi-
tion and velocity on a trajectory can be found in terms of the position and velocity at
an initial time. These functions are needed in the orbit determination algorithms of
Lambert and Gauss presented in Chapter 5.
The chapter concludes with a discussion of the restricted three-body problem in
order to provide a basis for understanding of the concepts of Lagrange points as well
as the Jacobi constant. This material is optional.
In studying this chapter it would be well from time to time to review the road
map provided in Appendix B.
2.2 Equations of motion in an inertial frame
Figure 2.1 shows two point masses acted upon only by the mutual force of gravity
between them. The positions of their centers of mass are shown relative to an inertial
frame of reference XYZ . The origin O of the frame may move with constant velocity
(relative to the fixed stars), but the axes do not rotate. Each of the two bodies is acted
upon by the gravitational attraction of the other. F12 is the force exerted on m1 by m2,
and F21 is the force exerted on m2 by m1.
The position vector RG of the center of mass G of the system in Figure 2.1(a) is,
defined by the formula
RG = m1R1 + m2R2
m1 + m2 (2.1)
X
Y
Z
O
r
Inertial frame of reference
(fixed with respect to the fixed stars)
GR1
R2
RG m2
m1
 
r
r
X
Y
Z
O
R1
R2
m2
m1
F12
F21
(a) (b)
ur �ˆ
Figure 2.1 (a) Two masses located in an inertial frame. (b) Free-body diagrams.
2.2 Equations of motion in an inertial frame 35
Therefore, the absolute velocity and the absolute acceleration of G are
vG = ṘG = m1Ṙ1 + m2Ṙ2
m1 + m2 (2.2)
aG = R̈G = m1R̈1 + m2R̈2
m1 + m2 (2.3)
The adjective ‘absolute’ means that the quantities are measured relative to an inertial
frame of reference.
Let r be the position vector of m2 relative to m1. Then
r = R2 − R1 (2.4)
Furthermore, let ûr be the unit vector pointing from m1 towards m2, so that
ûr = r
r
(2.5)
where r = ‖r‖, the magnitude of r. The body m1 is acted upon only by the force of
gravitational attraction towards m2. The force of gravitational attraction, Fg , which
acts along the line joining the centers of mass of m1 and m2, is given by Equation 1.3.
The force exerted on m2 by m1 is
F21 = Gm1m2
r2
(−ûr) = −Gm1m2
r2
ûr (2.6)
where −ûr accounts for the fact that the force vector F21 is directed from m2 towards
m1. (Do not confuse the symbol G, used in this context to represent the universal
gravitational constant, with its use elsewhere in the book to denote the center of
mass.) Newton’s second law of motion as applied to body m2 is F21 = m2R̈2, where
R̈2 is the absolute acceleration of m2. Thus
−Gm1m2
r2
ûr = m2R̈2 (2.7)
By Newton’s third law (the action–reaction principle), F12 = −F21, so that for m1 we
have
Gm1m2
r2
ûr = m1R̈1 (2.8)
Equations 2.7 and 2.8 are the equations of motion of the two bodies in inertial space.
By adding each side of these equations together, we find m1R̈1 + m2R̈2 = 0. According
to Equation 2.3, that means the acceleration of the center of mass G of the system of
two bodies m1 and m2 is zero. G moves with a constant velocity vG in a straight line,
so that its position vector relative to XYZ given by
RG = RG0 + vGt (2.9)
where RG0 is the position of G at time t = 0. The center of mass of a two-body system
may therefore serve as the origin of an inertial frame.
louiscoo
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louiscoo
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36 Chapter 2 The two-body problem
Example
2.1
Use the equations of motion to show why orbiting astronauts experience
weightlessness.
We sense weight by feeling the contact forces that develop wherever our body is
supported. Consider an astronaut of mass mA strapped into the space shuttle of mass
mS, in orbit about the earth. The distance between the center of the earth and the
spacecraft is r, and the mass of the earth is ME . Since the only external force on the
space shuttle is that of gravity, FS)g , the equation of motion of the shuttle is
FS)g = mSaS (a)
According to Equation 2.6,
FS)g = −GMEmS
r2
ûr (b)
where ûr is the unit vector pointing outward from the earth to the orbiting space
shuttle. Thus, (a) and (b) imply
aS = −GME
r2
ûr (c)
The equation of motion of the astronaut is
FA)g + CA = mAaA (d)
where FA)g is the force of gravity on (i.e., the weight of) the astronaut, CA is the
net contact force on the astronaut from restraints (e.g., seat, seat belt), and aA is the
astronaut’s acceleration. According to Equation 2.6,
FA)g = −GMEmA
r2
ûr (e)
Since the astronaut is moving with the shuttle we have, noting (c),
aA = aS = −GME
r2
ûr (f)
Substituting (e) and (f) into (d) yields
−GMEmA
r2
ûr + CA = mA
(
−GME
r2
ûr
)
from which it is clear that CA = 0. The net contact force on the astronaut is zero. With
no reaction to the force of gravity exerted on the body, there is no sensation of weight.
The potential energy V of the gravitational force in Equation 2.6 is given by
V = −Gm1m2
r
(2.10)
A force can be obtained from its potential energy function by means of the gradient
operator,
F = −∇V (2.11)
2.3 Equations of relative motion 37
where, in cartesian coordinates,
∇ = ∂
∂x
î + ∂
∂y
ĵ + ∂
∂z
k̂ (2.12)
In Appendix E it is shown that the gravitational potential, and hence the gravitational
force, outside of a sphere with a spherically symmetric mass distribution M is the
same as that of a point mass M located at the center of the sphere. Therefore, the two-
body problem applies not just to point masses but also to spherical bodies (as long,
of course, as they do not come into contact!).
2.3 Equations of relative motion
Let us now multiply Equation 2.7 by m1 and Equation 2.8 by m2 to obtain
−Gm
2
1m2
r2
ûr = m1m2R̈2
Gm1m22
r2
ûr = m1m2R̈1
Subtracting the second of these two equationsfrom the first yields
m1m2
(
R̈2 − R̈1
) = −Gm1m2
r2
(
m1 + m2
)
ûr
Canceling the common factor m1m2 and using Equation 2.4 yields
r̈ = −G(m1 + m2)
r2
ûr (2.13)
Let the gravitational µ parameter be defined as
µ = G(m1 + m2) (2.14)
The units of µ are km3s−2. Using Equation 2.14 together with Equation 2.5, we can
write Equation 2.13 as
r̈ = − µ
r3
r (2.15)
This is the second order differential equation that governs the motion of m2 relative to
m1. It has two vector constants of integration, each having three scalar components.
Therefore, Equation 2.15 has six constants of integration. Note that interchanging
the roles of m1 and m2 in all of the above amounts to simply multiplying Equation
2.15 through by −1, which, of course, changes nothing. Thus, the motion of m2 as
seen from m1 is precisely the same as the motion of m1 as seen from m2.
The relative position vector r in Equation 2.15 was defined in the inertial frame
(Equation 2.4). It is convenient, however, to measure the components of r in a frame
of reference attached to and moving with m1. In a co-moving reference frame, such
as the xyz system illustrated in Figure 2.2, r has the expression
r = xî + y ĵ + zk̂
38 Chapter 2 The two-body problem
X
Y
Z
O
R1 m2
m1
x
y
z
r
R2
ˆ i 
ˆ j 
ˆ k
Figure 2.2 Moving reference frame xyz attached to the center of mass of m1.
The relative velocity ṙrel and acceleration r̈rel in the co-moving frame are found by
simply taking the derivatives of the coefficients of the unit vectors, which themselves
are fixed in the xyz system. Thus
ṙrel = ẋî + ẏ ĵ + żk̂ r̈rel = ẍî + ÿ ĵ + z̈k̂
From Equation 1.40 we know that the relationship between absolute acceleration r̈
and relative acceleration r̈rel is
r̈ = r̈rel + �̇ × r + � × (� × r) + 2� × ṙrel
where � and �̇ are the angular velocity and angular acceleration of the moving frame
of reference. Thus r̈ = r̈rel only if � = �̇ = 0. That is to say, the relative acceleration
may be used on the left of Equation 2.15 as long as the co-moving frame in which it
is measured is not rotating.
As an example of two-body motion, consider two identical, isolated bodies m1
and m2 positioned in an inertial frame of reference, as shown in Figure 2.3. At time
t = 0, m1 is at rest at the origin of the frame, whereas m2, to the right of m1, has
a velocity vo directed upward to the right, making a 45◦ angle with the X axis. The
subsequent motion of the two bodies, which is due solely to their mutual gravitational
attraction, is determined relative to the inertial frame by means of Equations 2.7 and
2.8. Figure 2.3 is a computer-generated solution of those equations. The motion is
rather complex. Nevertheless, at any time t , m1 and m2 lie in the XY plane, equidistant
and in opposite directions from their center of mass G, whose straight-line path is
also shown in Figure 2.3. The very same motion appears rather less complex when
viewed from m1, as the computer simulation reveals in Figure 2.4(a). Figure 2.4(a)
2.3 Equations of relative motion 39
45°
vo
Gm1
(initially at rest)
m2
G
m1
m2
Path of G
Path of m1
Path of m2
X
Y
Inertial
frame
Figure 2.3 The motion of two identical bodies acted on only by their mutual gravitational attraction, as
viewed from the inertial frame of reference.
represents the solution to Equation 2.15, and we see that, relative to m1, m2 follows
what appears to be an elliptical path. (So does the center of mass.) Figure 2.4(b)
reveals that both m1 and m2 follow elliptical paths around the center of mass.
Since the center of mass has zero acceleration, we can use it as an inertial reference
frame. Let r1 and r2 be the position vectors of m1 and m2, respectively, relative to the
center of mass G in Figure 2.1. The equation of motion of m2 relative to the center of
mass is
−G m1m2
r2
ûr = m2r̈2 (2.16)
where, as before, r is the position vector of m2 relative to m1. In terms of r1 and r2,
r = r2 − r1
Since the position vector of the center of mass relative to itself is zero, it follows from
Equation 2.1 that
m1r1 + m2r2 = 0
Therefore,
r1 = −m2
m1
r2
so that
r = m1 + m2
m1
r2
Substituting this back into Equation 2.16 and using the fact that ûr = r2/r2, we get
−G m
3
1m2
(m1 + m2)2r32
r2 = m2r̈2
40 Chapter 2 The two-body problem
G
m2
m1
X
Y
Non-rotating frame
attached to m1
(a)
m1
X
Y
G
m2
Non-rotating frame
attached to G
(b)
Figure 2.4 The motion in Figure 2.3, (a) as viewed relative to m1 (or m2); (b) as viewed from the center
of mass.
which, upon simplification, becomes
−
(
m1
m1 + m2
)
µ
r32
r2 = r̈2 (2.17)
where µ is given by Equation 2.14. If we let
µ′ =
(
m1
m1 + m2
)3
µ
2.3 Equations of relative motion 41
vo
m1 m2 m3
m1
m1
m1
m2
m2
m3
m3
m2
G
X
Y
Inertial
frame
Figure 2.5 The motion of three identical masses as seen from the inertial frame in which m1 and m3 are
initially at rest, while m2 has an initial velocity v0 directed upwards and to the right, as shown.
then Equation 2.17 reduces to
r̈2 = −µ
′
r32
r2
which is identical in form to Equation 2.15.
In a similar fashion, the equation of motion of m1 relative to the center of mass is
found to be
r̈1 = −µ
′′
r31
r1
in which
µ′′ =
(
m2
m1 + m2
)3
µ
Since the equations of motion of either particle relative to the center of mass have the
same form as the equations of motion relative to either one of the bodies, m1 or m2,
it follows that the relative motion as viewed from these different perspectives must be
similar, as illustrated in Figure 2.4.
One may wonder what the motion looks like if there are more than two bodies
moving under the influence only of their mutual gravitational attraction. The n-body
problem with n > 2 has no closed form solution, which is complex and chaotic in
nature. We can use a computer simulation (see Appendix C.1) to get an idea of the
motion for some special cases. Figure 2.5 shows the motion of three equal masses,
42 Chapter 2 The two-body problem
m1
m1
m1
m2
m2
m3
m3
m3
X
Y
Non-rotating frame
attached to G
Figure 2.6 The same motion as Figure 2.5, as viewed from the inertial frame attached to the center of
mass G.
equally spaced initially along the X axis of an inertial frame. The center mass has an
initial velocity, while the other two are at rest. As time progresses, we see no periodic
behavior as was evident in the two-body motion in Figure 2.3. The chaos is more
obvious if the motion is viewed from the center of mass of the three-body system, as
shown in Figure 2.6. The computer simulation from which these figures were taken
shows that the masses eventually collide.
2.4 Angular momentum and the orbit formulas
The angular momentum of body m2 relative to m1 is the moment of m2’s relative
linear momentum m2ṙ (cf. Equation 1.17),
H2/1 = r × m2ṙ
where ṙ = v is the velocity of m2 relative to m1. Let us divide this equation through
by m2 and let h = H2/1/m2, so that
h = r × ṙ (2.18)
h is the relative angular momentum of m2 per unit mass, that is, the specific relative
angular momentum. The units of h are km2 s−1.
Taking the time derivative of h yields
dh
dt
= ṙ × ṙ + r × r̈
But ṙ × ṙ = 0. Furthermore, r̈ = −(µ/r3)r, according to Equation 2.15, so that
r × r̈ = r ×
(
− µ
r3
r
)
= − µ
r3
(r × r) = 0
2.4 Angular momentum and the orbit formulas 43
m1
m2
r
r
·r
·r
ĥ �
h
h
ĥ �
h
h
Figure 2.7 The path of m2 around m1 lies in a plane whose normal is defined by h.
m1 m2
r
û⊥
û r
υr
υ⊥
Path
r·
Figure 2.8 Components of the velocity of m2, viewed above the plane of the orbit.
Therefore,
dh
dt
= 0 (or r × ṙ = constant) (2.19)
At any given time, the position vector r and the velocity vector ṙ lie in the same plane,
as illustrated in Figure 2.7. Their cross product r × ṙ is perpendicular to that plane.
Since r × ṙ = h, the unit vector normal to the plane is
ĥ = h
h
(2.20)
But, according to Equation 2.19, this unit vector is constant. Thus, the path of m2
around m1 lies in a single plane.
Since the